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					       Lecture 10
     Potential Energy

 Potential   Energy
  Definition
  Conservative Forces


 Conservation   of Energy

 Power
    Work Done by Gravity
 Wg   = -mgy
  Independent of path (only depends on
   initial and final height)
  If you end up where you began, Wg = 0
  Define: Potential Energy: Ug = mgy
  Thus: Ug = mgy = -Wg



 We call gravity a “Conservative Force” because Wg
is path independent (and therefore we can define a
          “Potential Energy” to go with it).
     Potential (stored) Energy
   Works for any CONSERVATIVE force
    Gravity Ug = m g y
    Spring Us = ½ k x2
    NOT friction


   “Stored” Gravitational Energy can be
    converted to Kinetic Energy
    Wg= K
    -m g y = K
    0 = K + m g y
    0 = K + Ug
Work - Energy w/ Conservative Forces

           Work-Energy Theorem:
                 W  K
 Move work by conservative forces to other side:
               Wnc  K  U
    If there are NO non-conservative forces:
                 0  K  U
             Ki  U i  K f  U f
      Power (Rate of Work)
P   = W / t
  Units: Joules/Second = Watt (W)


 But   Remember:
  W = F x cosq = F (v t) cosq
  P = F v cosq
           Summary
Conservative Forces
 » Work is independent of path
 » Define Potential Energy U
    Ugravity = m g y

    Uspring = ½ k x2



Work – Energy Theorem

        Wnc  K  U
                         Example
     Standing at the top of a cliff you throw a ball at
      12 m/s. If the cliff is 87 m high, how fast is the
      ball moving right before it hits the ground?




We will use the Work-Energy Theorem:


        Wnc  K  U
                       Example
   Standing at the top of a cliff you throw a ball at
    12 m/s. If the cliff is 87 m high, how fast is the
    ball moving right before it hits the ground?




       0  K  U
     Ki  U i  K f  U f
                        Example
    Standing at the top of a cliff you throw a ball at
     12 m/s. If the cliff is 87 m high, how fast is the
     ball moving right before it hits the ground?


1 2           1 2
  mvi  mgyi  mv f  mgy f
2             2
     cancel m from each expression
     vi = 12 m/s
     yi = 87 m
     yf = 0 m
     solve for vf
                             Example
     Standing at the top of a cliff you throw a ball at
      12 m/s. If the cliff is 87 m high, how fast is the
      ball moving right before it hits the ground?


                Vf = 43 m/s

Notes:

  The final speed did not depend on the angle the
   ball was thrown upward (although the angle
   would affect how quickly the ball hits the ground).

  The mass of the ball did not affect the final speed.

				
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posted:8/5/2012
language:English
pages:10